The value of c for which the Mean Value Theor | vedclass

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(D) Given function is $f(x) = \log_{e} x$ on the interval $[1, 3]$.First,we calculate the values at the endpoints:$f(1) = \log_{e} 1 = 0$$f(3) = \log_

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$2 \log_{3} e$

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(D) Given function is $f(x) = \log_{e} x$ on the interval $[1, 3]$.First,we calculate the values at the endpoints:$f(1) = \log_{e} 1 = 0$$f(3) = \log_{e} 3$Next,we find the derivative of the function:$f'(x) = \frac{1}{x}$According to Lagrange's Mean Value Theorem,there exists a point $c \in (1, 3)$ such that:$f'(c) = \frac{f(3) - f(1)}{3 - 1}$Substituting the values:$\frac{1}{c} = \frac{\log_{e} 3 - 0}{2}$$\frac{1}{c} = \frac{\log_{e} 3}{2}$$c = \frac{2}{\log_{e} 3}$Using the property of logarithms $\frac{1}{\log_{a} b} = \log_{b} a$,we get:$c = 2 \log_{3} e$Thus,the correct option is $D$.

The value of $c$ for which the Mean Value Theorem holds for the function $f(x) = \log_{e} x$ on the interval $[1, 3]$ is:

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